## Algebra Solutions: Hungerford I.4

1) Let G be a group and $\{H_i \mid i\in I\}$ a family of subgroups. Then for any $a\in G \text{, } (\cap_i H_i)a = \cap_i H_ia$.
Proof-
$(\cap_i H_i)a = \{x\in G \mid x \equiv_r a \pmod{\cap_i H_i}\} = \{ x\in G \mid xa^{-1} \in \cap_i H_i \}= \{ x\in G \mid xa^{-1} \in H_i \forall i \} = \{ x\in G \mid x \equiv_r a \pmod{H_i} \forall i \} = \{ x\in G \mid x \in H_ia \forall i \} = \cap_i H_ia$

3) The following conditions on a finite group G are equivalent.
(i) |G| is prime
(ii) $G \neq \langle e \rangle$ and G has no proper subgroups.
(iii) $G \cong \mathbb{Z}_p$ for some prime p.
Proof-
$(i) \Rightarrow (iii)$ Assume that the order of G is a prime, p. Then G has at least 2 elements. Pick an element a, which is not the identity and consider $\langle a \rangle$. By Lagrange, we know the order of a divides the order of the group. Thus the order of a is either 1 or p, but it is not 1 since a is not the identity. Thus a generates G, so G is cyclic of order p and hence isomorphic to $\mathbb{Z}_p$ since any finite cyclic group of a certain order is actually the same.
$(iii) \Rightarrow (ii)$ Assume that $G \cong \mathbb{Z}_p$ for some prime p. Then the order of G is p which is greater than or equal to 2, so G is nontrivial. Let $H \leq G$. By Lagrange again, the order of H divides the order of G, so that the order of H is either 1 or p. The first case gives H to be trivial, while the second gives H to be the entire group. Thus G has no proper subgroups.
$(ii) \Rightarrow (iii)$ Assume that G is not trivial and G has no proper subgroups. Since G is nontrivial, there is some a in G such that a is not the identity. Since G has no proper subgroups, $\langle a \rangle = G$ so that G is cyclic and of finite order, hence isomorphic to $\mathbb{Z}_m$ for some integer m. Suppose m is not prime then there is some natural number k such that k divides m and k is not 1 or m. Then because G is cyclic and k divides the order of G, there is exactly 1 subgroup of order k. However we know G has no proper subgroups, so this is a contradiction. Therefore m is prime.
$(iii) \Rightarrow (i)$ Trivial

4) Let $a \in \mathbb{Z}$ and p be a prime such that $p \nmid a$. Then $a^{p-1} \equiv 1 \pmod{p}$.
Proof-
By the division algorithm $a = pq + r$ for some $q,r \in \mathbb{Z}$ with $0 < r < p$. Thus taking the mod of both sides by p we get $a \equiv r \pmod{p}$. Recall that the nonzero elements of $\mathbb{Z}_p$ form a group of order p-1 under multiplication mod p. Since $0 r is in this group. Therefore the order of r, call it n, must divide p-1, the order of the group. Thus $p-1 = nm$ for some integer m. So we have:
$a^{p-1} \equiv r^{p-1} \pmod{p}$ but
$r^{p-1} = r^{nm} = (r^{n})^m = 1^m = 1$ so that
$a^{p-1} \equiv 1 \pmod{p}$

6) Let H, K be subgroups of a group G. Then $HK \leq H \text{ iff } HK = KH$
Proof-
$( \Rightarrow )$ Assume that $HK \leq G$. We prove that $HK = KH$ by subset inclusion. Let $g\in HK$ Then g can be written $g = hk$ for some $h \in H \text{, } k \in K$. Since $HK \leq G \text{ then } g^{-1} \in HK$ as well. Note, $g^{-1} = (hk)^{-1} = k^{-1}h^{-1}$ so that we know $k^{-1}h^{-1} \in HK$. This tells us that there is some $h' \in H \text{ and } k' \in K$ such that $k^{-1}h^{-1} = k'h'$. Since H and K are subgroups of G, $(h')^{-1} \in H \text{ and } (k')^{-1} \in K$. Putting everything together we have that:
$g = (g^{-1})^{-1} = (k^{-1}h^{-1})^{-1} = (k'h')^{-1} = (h')^{-1}(k')^{-1} \in KH$.
Now let $g \in KH$ so that $g=kh$ for some $k \in K \text{ and } h \in H$. Then we have that $g^{-1} = (kh)^{-1} = h^{-1} k^{-1}$. Since H and K are subgroups we have $h^{-1} \in H$ and $k^{-1} \in K$, so that $g^{-1} = h^{-1}k^{-1} \in HK$ and HK is a subgroup so that $g = (g^{-1})^{-1} \in HK$ also. Thus HK = KH
$( \Leftarrow )$ Assume now that HK = KH and show $HK \leq G$. Since H and K are subgroups of G, we have that the $e \in H \cap K$ so that $e = ee \in HK$ which makes HK nonempty. Now we use the subgroup test:
Let $hk,h'k' \in HK$ Then $(hk)(h'k')^{-1} = hk(k')^{-1}(h')^{-1}$ Note that $k(k')^{-1}(h')^{-1} \in KH = HK$ so that we can write $k(k')^{-1}(h')^{-1} = h''k''$ Then we have:
$(hk)(h'k')^{-1} = hh''k'' \in HK$. So HK is a subgroup of G.

7) Let G be a group s.t. $|G| = p^km$ where p is prime and gcd(p,m) = 1. Let H, K be supgroups of G s.t. $|H| = p^k \text{ and } |K| = p^d$ with $0 < d \leq k$ and K not a subgroup of H. Show that $HK \nleq G$
Proof-
(By contradiction) Assume that $HK \leq G$ then we get $|HK| = \frac{|H||K|}{|H \cap K|} = \frac{p^kp^d}{|H \cap K|}$. We know that the intersection of subgroups is a subgroup so $H \cap K \leq G$ so that in order for $|HK| \in \mathbb{N}$ it must be the case that $|H\cap k| = p^s$ for some $s \leq k+d$. However since K is not a subgroup of H, the intersection of H and K must contain strictly less elements than K. Thus:
$p^s = |H\cap K| < |K| = p^d$ so that s < d. Now we have that $|HK| = p^{k+d-s}$. By Lagrange the order of the subgroup must divide the order of G, so that $p^{k+d-s} \mid p^km$ which is the same as saying, for some $n \in \mathbb{N}$ that:
$p^km = p^{k+d-s}n \Rightarrow m = p^{d-s}n \Rightarrow m = p(p^{d-s-1}n) \Rightarrow p \mid m$ This is a contradiction because gcd(p,m) = 1. Thus HK is not a subgroup of G.

8) If H and K are subgroups of finite index of a group G such that $[G:H] \text{ and } [G:K]$ are relatively prime, then G = HK.
Proof-
First note two facts: $H\cap K \leq H \leq G$ and $H\cap K \leq K \leq G$ so by theorem 4.5 we can say:
$[G: H\cap K] = [G: K] [K: H\cap K]$ and
$[G: H\cap K] = [G: H] [H: H\cap K]$. This tells us that $[G: H\cap K]$ is a common multiple of both $[G:H] \text{ and } [G:K]$. However because $gcd([G:H],[G:K]) = 1$ we have that $lcm([G:H],[G:K]) = [G:H][G:K]$. Then clearly because $[G: H\cap K]$ is a common multiple it must be larger than or equal to the least common multiple so that $[G: H\cap K] \geq [G:H][G:K]$. Lastly, by proposition 4.9 we have $[G: H\cap K] \leq [G:H][G:K]$. Therefore $[G: H\cap K] = [G:H][G:K]$ so again by proposition 4.9 we can say that G = HK.

9) If H, K, and N are subgroups of a group G such that $H\leq N$, then $HK \cap N = H ( K\cap N)$
Proof-
By subset inclusion. Let $x\in HK \cap N$ Then we have $x\in HK \text{ and } x\in N$. From this we can say $x= hk$ for some $h\in H , k\in K$. So $k = h^{-1}x$ Since H is given as a subgroup of N this means $h^{-1} \in N$. so that $k = h^{-1}x \in N$ by closure. But $k\in K$ so that $k\in K \cap N$ Thus:
$x = hk\in H(K\cap N)$ since $h\in H, k \in K cap N$.
Now let $x \in H(K\cap N)$ This says that $x =hg$ where $h\in H \text{ and } g\in K\cap N$. Since this tells us g is in K we have $x\in HK$. Note H is a subgroup of N so that h is in N, and also g is in N. Thus their product is in N. So in conclusion we have $x\in HK \cap N$

10) Let H,K,N be subgroups of a group G such that $H \leq K, H\cap N = K\cap N, \text{ and } HN =KN$. Show that H = K.
Proof-
H is already a subgroup of K so we need only show that K is inside of H. Let $k\in K$. Then k can be written as $k = ke \in KN = HN$ which says that for some $h\in H \text{ and } n\in N$ we have $k = hn$. Then $h^{-1}k = n \in N$. Also we have $h^{-1} \in K$because H is a subgroup of K. Therefore we have $h^{-1}k \in N \text{ and } h^{-1}k \in K$ so that$h^{-1}k \in N\cap K = H\cap N$.Note the intersection of H and N is inside of H, so that $h^{-1}k \in H$. Thus $h^{-1}k = h' \Rightarrow k = hh' \in H$. So H = K.

11) Let G be a group of order 2n, then G contains an element of order 2. If n is odd and G is abelian, there is only one element of order 2.
Proof-
Let the order of G be 2n. Consider the set $\{e, a_1,a_1^{-1}, a_2,a_2^{-1}, ..., a_m,a_m^{-1}\}$. If there does not exist an element $a_i$ such that $a_i^{-1} = a_i$ then the order of this set is $2m+1$, which is a contradiction so there must exist at least one element of order 2.
If G is abelian and n is odd, suppose we have two elements $a,b \in G$ such that $a^2 = e = b^2$. G is abelian so that $(ab)^2 = e \Rightarrow \{e,a,b,ab\}$ is a subgroup of G. But because n is odd, we know that 4 does not divide 2n, so this subgroup cannot exist by Lagrange. Thus there can be only one element of order 2.